Problem: $\dfrac{ -9x - 6y }{ 9 } = \dfrac{ x - 3z }{ -3 }$ Solve for $x$.
Answer: Multiply both sides by the left denominator. $\dfrac{ -9x - 6y }{ {9} } = \dfrac{ x - 3z }{ -3 }$ ${9} \cdot \dfrac{ -9x - 6y }{ {9} } = {9} \cdot \dfrac{ x - 3z }{ -3 }$ $-9x - 6y = {9} \cdot \dfrac { x - 3z }{ -3 }$ Reduce the right side. $-9x - 6y = {9} \cdot \dfrac{ x - 3z }{ -{3} }$ $-9x - 6y = -{3} \cdot \left( x - 3z \right)$ Distribute the right side $-9x - 6y = -{3} \cdot \left( {x} - {3z} \right)$ $-9x - 6y = -{3}x + {9}z$ Combine $x$ terms on the left. $-{9x} - 6y = -{3x} + 9z$ $-{6x} - 6y = 9z$ Move the $y$ term to the right. $-6x - {6y} = 9z$ $-6x = 9z + {6y}$ Isolate $x$ by dividing both sides by its coefficient. $-{6}x = 9z + 6y$ $x = \dfrac{ 9z + 6y }{ -{6} }$ All of these terms are divisible by $3$ Divide by the common factor and swap signs so the denominator isn't negative. $x = \dfrac{ -{3}z - {2}y }{ {2} }$